What are the solutions to #(x-3)^2+8=12#?

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Answer 1 · 2018-04-16T15:52:56

#x=5##" or "##x=1#

#(a+b)^2=a^2+2ab+b^2#\

Use the same rule

#(x-3)^2=x^2+2(x)(-3)+(-3)^2#
#=x^2-6x+9#

now Substitute

#x^2-6x+9+8=12#

#x^2-6x+5=0#

Factorize

#(x-5)(x-1)=0#

#x=5##" "##x=1#